Trouble understanding semidirect product of $Z_3$ and $Z_{13}$

Question

Here a representation for a non Abelian of order 39 is given
Finding presentation of group of order 39 by Pisco. Am I correct in understanding that the non-trivial homomorphism in this case is:
$\phi:Z_3 \rightarrow Aut(Z_{13})$ where $\phi_y(x) = yxy^{-1}$ $\forall y\in Z_3$ and $\forall x\in Z_{13}$
For different $y$ we get different Automorphisms in $Z_{13}$.
To find the actual automorphism i.e. to find $i$ s.t. $yxy^{-1} = x^i$, we are using the automorphism corresponding to the generator $y$ of $Z_3$ and $x$ of $Z_{13}$? Appreciate your help in explaining the semidirect product in terms of its explicit definition to help me get used to the idea.

Answer 1

Your homomorphism $\phi$ makes no sense to me; in $$\phi:\ \Bbb{Z}/3\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z}/13\Bbb{Z}):\ y\ \longmapsto\ (x\ \longmapsto\ yxy^{-1}),$$ how do you multiply elements $y\in\Bbb{Z}/3\Bbb{Z}$ and $x\in\Bbb{Z}/13\Bbb{Z}$ together? Let me give some background.

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In stead, identify $\operatorname{Aut}(\Bbb{Z}/13\Bbb{Z})\cong(\Bbb{Z}/13\Bbb{Z})^{\times}$ with $\Bbb{Z}/12\Bbb{Z}$ by choosing a primitive root mod $13$, i.e. a generator of the multiplicative group $(\Bbb{Z}/13\Bbb{Z})^{\times}$.

For example, a few calculations show that $2$ is a primitive root mod $13$. Hence every automorphism of $\Bbb{Z}/13\Bbb{Z}$ is determined by where it sends $2$. So every automorphism is of the form $$\psi:\ \Bbb{Z}/13\Bbb{Z}\ \longmapsto\ \Bbb{Z}/13\Bbb{Z}:\ 2\ \longmapsto\ 2^k,$$ for some $k\in\{1,\ldots,12\}$. Check that this gives an isomorphism $$\Bbb{Z}/12\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z}/13\Bbb{Z}):\ k\ \longmapsto\ (n\ \longmapsto\ n\cdot2^k).$$

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This makes it clearer what homomorphisms $\phi:\ \Bbb{Z}/3\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z}/13\Bbb{Z})$ look like. They are determined by where $1\in\Bbb{Z}/3\Bbb{Z}$ is mapped, and this must be mapped to an element of order dividing $3$. There are only three elements of order dividing $3$ in $\Bbb{Z}/12\Bbb{Z}$, they are $0$, $4$ and $8$. So we have three homomorphisms $$\phi:\ \Bbb{Z}/3\Bbb{Z}\ \longrightarrow\ \operatorname{Aut}(\Bbb{Z}/13\Bbb{Z}),$$ and they are of the form $$\phi(m):\ \Bbb{Z}/13\Bbb{Z}\ \longrightarrow\ \Bbb{Z}/13\Bbb{Z}:\ n\ \longmapsto\ n\cdot2^{km},$$ for some $k\in\{0,4,8\}$.